# I want to do exponential fitting for power decaying with time

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Hozifa on 21 Sep 2022
Answered: Image Analyst on 21 Sep 2022
Hi there,
I have this matrix (first cloumn is time while the second column is received power), I want to do exponentially fitting, how can I can do it and how can I find time-decaying constant of the exponential.
x=[
65.10 1.0000e+000
65.90 61.0843e-003
70.30 21.3693e-003
];

Image Analyst on 21 Sep 2022
If you have the Statistics and Machine Learning Toolbox you can use fitnlm. See attached full demo. Replace the demo data with your own. However I'll tell you that any fit with only 3 data points will probably be very inaccurate. You should definitely make more measurements.

Torsten on 21 Sep 2022
x=[
65.10 1.0000e+000
65.90 61.0843e-003
70.30 21.3693e-003
];
fun = @(p)exp(-p*(x(:,1)-65.1));
fun1 = @(p)fun(p)-x(:,2);
p = lsqnonlin(fun1,0.1)
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
p = 3.4943
format long
fun1(p)
ans = 3×1
0 0.000002202330315 -0.021369287157659
hold on
plot(x(:,1),x(:,2))
plot(x(:,1),fun(p))
hold off

Cris LaPierre on 21 Sep 2022
The easiest approach is to use the Curve Fitting app inside a live script. This does require having the Curve Fitting Toolbox installed. Once done, you can have the app generate the corresponding code. Here is what that code might look like.
x=[
65.10 1.0000e+000
65.90 61.0843e-003
70.30 21.3693e-003
];
X = x(:,1);
Y = x(:,2);
[xData, yData] = prepareCurveData( X, Y );
% Set up fittype and options.
ft = fittype( 'exp1' );
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
opts.Display = 'Off';
opts.StartPoint = [1.19913731842403e+35 -1.24044938669103];
% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, opts );
% Plot fit with data.
figure( 'Name', 'untitled fit 1' );
h = plot( fitresult, xData, yData );
legend( h, 'Y vs. X', 'untitled fit 1', 'Location', 'NorthEast', 'Interpreter', 'none' );
% Label axes
xlabel( 'X', 'Interpreter', 'none' );
ylabel( 'Y', 'Interpreter', 'none' );
grid on
I assume you are intersted in the coefficient b.
fitresult
fitresult =
General model Exp1: fitresult(x) = a*exp(b*x) Coefficients (with 95% confidence bounds): a = 6.236e+98 (-2.254e+101, 2.266e+101) b = -3.494 (-9.061, 2.072)